Let S be the set of all real numbers. Show that the relation R = {(a,

1.	Let S be the set of all real numbers. Show that the relation R = {(a, b): a2 + b2 = 1} is symmetric but neither reflexive nor transitive.
Answer» We know that R = {(a, b): a² + b²= 1} Reflexive- R is the set of real numbers If a ∈ R then a² + a²≠ 1 where a = 2, 3 …. Therefore, R is not reflexive. Symmetric- Consider (a, b) ∈ R where a² + b²= 1 We get (b, a) ∈ R and (a, b) ∈ R Hence, R is symmetric. Transitive- Consider (a, b) ∈ R and (b, c) ∈ R We know that (cos 30^o, sin 30^o) ∈ R and (sin 30^o, cos 30^o) ∈ R So (cos 30^o, cos 30^o) ∉ R Therefore, R is not transitive. Here, R is symmetric but neither reflexive nor transitive.

Let S be the set of all real numbers. Show that the relation R = {(a, b): a2 + b2 = 1} is symmetric but neither reflexive nor transitive.

Answer»

We know that

R = {(a, b): a² + b²= 1}

Reflexive-

R is the set of real numbers

If a ∈ R then a² + a²≠ 1 where a = 2, 3 ….

Therefore, R is not reflexive.

Symmetric-

Consider (a, b) ∈ R where a² + b²= 1

We get

(b, a) ∈ R and (a, b) ∈ R

Hence, R is symmetric.

Transitive-

Consider (a, b) ∈ R and (b, c) ∈ R

We know that

(cos 30^o, sin 30^o) ∈ R and (sin 30^o, cos 30^o) ∈ R

So (cos 30^o, cos 30^o) ∉ R

Therefore, R is not transitive.

Here, R is symmetric but neither reflexive nor transitive.

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